Question

In an equilateral triangle ABC, D is the midpoint of side BC. Prove that 4AD${}^2$ = 3BC${}^2$.

Answer 1

Let the side of triangle be $BC=a$

The height of Triangle $AD=\frac{\sqrt{3}}{2}a$

$4A{D}^2=4{\left(\frac{\sqrt{3}a}{2}\right)}^{2}4\left(\frac{3}{4}{a}^2\right)=3a^2=3B{C}^2$